Question 911581
I'm assuming these are consecutive even or odd numbers since consecutive numbers will not yield an integer answer.
{{{3(n-2)n(n+2)=5(n-2+n+n+2)}}}
{{{3n(n^2-4)=15n}}}
{{{3n^3-12n=15n}}}
{{{3n^3-27n=0}}}
{{{n^3-9n=0}}}
{{{n(n^2-9)=0}}}
{{{n(n-3)(n+3)=0}}}
Three solutions:
{{{n=0}}} Then the numbers are -2,0, and 2.
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{{{n-3=0}}}
{{{n=3}}} Then the numbers are 1,3, and 5.
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{{{n+3=0}}}
{{{n=-3}}} Then the numbers are -5,-3, and -1.